(3x-4)^2-6x=7(1-x)(1+x)

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Solution for (3x-4)^2-6x=7(1-x)(1+x) equation: 



(3x-4)^2-6x=7(1-x)(1+x)

We move all terms to the left:

(3x-4)^2-6x-(7(1-x)(1+x))=0

We add all the numbers together, and all the variables

(3x-4)^2-6x-(7(-1x+1)(x+1))=0

We add all the numbers together, and all the variables

-6x+(3x-4)^2-(7(-1x+1)(x+1))=0

We multiply parentheses ..

-(7(-1x^2-1x+x+1))-6x+(3x-4)^2=0



We calculate terms in parentheses: -(7(-1x^2-1x+x+1)), so:

7(-1x^2-1x+x+1)

We multiply parentheses

-7x^2-7x+7x+7

We add all the numbers together, and all the variables

-7x^2+7

Back to the equation:

-(-7x^2+7)



We get rid of parentheses

7x^2-6x+(3x-4)^2-7=0

We move all terms containing x to the left, all other terms to the right

7x^2-6x+(3x-4)^2=7

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